Let *y = m _{1}x + c_{1}* and

*y = m*

_{2}x + c_{2 }be the equations of two lines in a plane where,

m_{1}= slope of line 1 c_{1}= y-intercept made by line 1 m2 = slope of line 2 c2 = y-intercept made by line 2

<BAX = θ_{1}<DCX = θ_{2 }∴ m_{1 }= tan θ_{1 }and m_{2 }= tan θ_{2}

Also if we consider *<APD* as the angle between lines,

Hence, the angle between two lines is,

**Condition for perpendicularity**

The two lines are perpendicular means. Ø = 90°

Thus, the lines are perpendicular if the product of their slope is -1.

**Condition for parallelism**

The two lines are perpendicular means, Ø = 0°

Thus, the lines are parallel if their slopes are equal.

### Angle Between Two Lines Examples

1. Find the angle between the lines *2x-3y+7 = 0* and *7x+4y-9 = 0*.

**Solution:**

Comparing the equation with equation of straight line, *y = mx + c*,

Slope of line 2x-3y+7=0 is (m_{1}) = 2/3

Slope of line 7x+4y-9=0 is (m_{2}) = -7/4

Let, *Ø* be the angle between two lines, then

2. Find the equation of line through point (3,2) and making angle 45*°* with the line *x-2y = 3*.

**Solution:**

Let *m* be the slope of the required line passing through (3,2). So, using slope point form, its equation is

*y-2 = m(x-3)* ——— (i)

Slope of line *x-2y = 3* is 1/2.

Since, these lines make an angle of 45*° *so,

Substituting values of m in equation (i), we get

*y-2 = 3(x-3)* and *y-2 = -(x-3)/3*

or, *3x-y-7 = 0* and *x+3y-9 = 0* are the required equations of line.